How much iron (3) hydroxide should be taken to react with sulfuric acid to obtain 86g of salt?

Let’s implement the solution:
1. In accordance with the condition of the problem, we write the equation:
X g -? m = 86 g;
2Fe (OH) 3 + 3H2SO4 = Fe2 (SO4) 3 + 6H2O – ion exchange reaction, ferrous sulfate, water is released;
2. We make calculations:
M Fe (OH) 3 = 106.8 g / mol;
M Fe2 (SO4) 3 = 399.6 g / mol.
3. Calculate the number of moles of iron sulfate, if the mass is known:
Y Fe2 (SO4) 3 = m / M = 86 / 399.6 = 0.215 mol.
4. Find the number of moles of the starting material:
X mol Fe (OH) 3 – 0.215 mol Fe2 (SO4) 3;
-2 mol -1 mol from here, X mol Fe (OH) 3 = 2 * 0.215 / 1 = 0.43 mol.
5. We calculate the mass of the initial substance:
m Fe (OH) 3 = Y * M = 0.43 * 106.8 = 45.92 g.
Answer: to carry out the process, you will need iron hydroxide weighing 45.92 g.